at1124canvas_practice
Use completing-the-square to solve the equation below.
\[x^2+14x~=~98\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+14x~=~98\]
The linear coefficient is 14. Square half of 14 to get 49. Add this amount to both sides to complete the square.
\[x^2+14x+49~=~98+49\]
Simplify the right side.
\[x^2+14x+49~=~147\]
Factor the left side.
\[(x+7)^2~=~147\]
Undo the squaring.
\[x+7~=~\pm \sqrt{147}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+7~=~\pm \sqrt{3^{1}\cdot7^{2}}\]
\[x+7~=~\pm 7\sqrt{3}\]
Subtract 7 from both sides.
\[x~=~-7\pm 7\sqrt{3}\]
So,
- \(m=-7\)
- \(n=7\)
- \(p=3\)
Use completing-the-square to solve the equation below.
\[x^2+8x~=~59\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+8x~=~59\]
The linear coefficient is 8. Square half of 8 to get 16. Add this amount to both sides to complete the square.
\[x^2+8x+16~=~59+16\]
Simplify the right side.
\[x^2+8x+16~=~75\]
Factor the left side.
\[(x+4)^2~=~75\]
Undo the squaring.
\[x+4~=~\pm \sqrt{75}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+4~=~\pm \sqrt{3^{1}\cdot5^{2}}\]
\[x+4~=~\pm 5\sqrt{3}\]
Subtract 4 from both sides.
\[x~=~-4\pm 5\sqrt{3}\]
So,
- \(m=-4\)
- \(n=5\)
- \(p=3\)
Use completing-the-square to solve the equation below.
\[x^2+14x~=~-9\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+14x~=~-9\]
The linear coefficient is 14. Square half of 14 to get 49. Add this amount to both sides to complete the square.
\[x^2+14x+49~=~-9+49\]
Simplify the right side.
\[x^2+14x+49~=~40\]
Factor the left side.
\[(x+7)^2~=~40\]
Undo the squaring.
\[x+7~=~\pm \sqrt{40}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+7~=~\pm \sqrt{2^{3}\cdot5^{1}}\]
\[x+7~=~\pm 2\sqrt{10}\]
Subtract 7 from both sides.
\[x~=~-7\pm 2\sqrt{10}\]
So,
- \(m=-7\)
- \(n=2\)
- \(p=10\)
Use completing-the-square to solve the equation below.
\[x^2-8x~=~82\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2-8x~=~82\]
The linear coefficient is -8. Square half of -8 to get 16. Add this amount to both sides to complete the square.
\[x^2-8x+16~=~82+16\]
Simplify the right side.
\[x^2-8x+16~=~98\]
Factor the left side.
\[(x-4)^2~=~98\]
Undo the squaring.
\[x-4~=~\pm \sqrt{98}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x-4~=~\pm \sqrt{2^{1}\cdot7^{2}}\]
\[x-4~=~\pm 7\sqrt{2}\]
Add 4 to both sides.
\[x~=~4\pm 7\sqrt{2}\]
So,
- \(m=4\)
- \(n=7\)
- \(p=2\)
Use completing-the-square to solve the equation below.
\[x^2-8x~=~92\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2-8x~=~92\]
The linear coefficient is -8. Square half of -8 to get 16. Add this amount to both sides to complete the square.
\[x^2-8x+16~=~92+16\]
Simplify the right side.
\[x^2-8x+16~=~108\]
Factor the left side.
\[(x-4)^2~=~108\]
Undo the squaring.
\[x-4~=~\pm \sqrt{108}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x-4~=~\pm \sqrt{2^{2}\cdot3^{3}}\]
\[x-4~=~\pm 6\sqrt{3}\]
Add 4 to both sides.
\[x~=~4\pm 6\sqrt{3}\]
So,
- \(m=4\)
- \(n=6\)
- \(p=3\)
Use completing-the-square to solve the equation below.
\[x^2+6x~=~9\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+6x~=~9\]
The linear coefficient is 6. Square half of 6 to get 9. Add this amount to both sides to complete the square.
\[x^2+6x+9~=~9+9\]
Simplify the right side.
\[x^2+6x+9~=~18\]
Factor the left side.
\[(x+3)^2~=~18\]
Undo the squaring.
\[x+3~=~\pm \sqrt{18}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+3~=~\pm \sqrt{2^{1}\cdot3^{2}}\]
\[x+3~=~\pm 3\sqrt{2}\]
Subtract 3 from both sides.
\[x~=~-3\pm 3\sqrt{2}\]
So,
- \(m=-3\)
- \(n=3\)
- \(p=2\)
Use completing-the-square to solve the equation below.
\[x^2+16x~=~-44\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+16x~=~-44\]
The linear coefficient is 16. Square half of 16 to get 64. Add this amount to both sides to complete the square.
\[x^2+16x+64~=~-44+64\]
Simplify the right side.
\[x^2+16x+64~=~20\]
Factor the left side.
\[(x+8)^2~=~20\]
Undo the squaring.
\[x+8~=~\pm \sqrt{20}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+8~=~\pm \sqrt{2^{2}\cdot5^{1}}\]
\[x+8~=~\pm 2\sqrt{5}\]
Subtract 8 from both sides.
\[x~=~-8\pm 2\sqrt{5}\]
So,
- \(m=-8\)
- \(n=2\)
- \(p=5\)
Use completing-the-square to solve the equation below.
\[x^2+10x~=~87\]
The solution, when simplified, can be expressed in the form shown below.
\[x~=~m \pm n\sqrt{p}\]
- \(m=\)
- \(n=\)
- \(p=\)
\[x^2+10x~=~87\]
The linear coefficient is 10. Square half of 10 to get 25. Add this amount to both sides to complete the square.
\[x^2+10x+25~=~87+25\]
Simplify the right side.
\[x^2+10x+25~=~112\]
Factor the left side.
\[(x+5)^2~=~112\]
Undo the squaring.
\[x+5~=~\pm \sqrt{112}\]
Simplify the radical. I recommend using prime factorization (with a factor tree).
\[x+5~=~\pm \sqrt{2^{4}\cdot7^{1}}\]
\[x+5~=~\pm 4\sqrt{7}\]
Subtract 5 from both sides.
\[x~=~-5\pm 4\sqrt{7}\]
So,
- \(m=-5\)
- \(n=4\)
- \(p=7\)
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}-{16}x-{70}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}+{16}x+{70}\right)\]
\[\left(\frac{16}{2}\right)^2~=~64\]
\[f(x)~=~-\left(x^{2}+{16}x+{64}+{70}-{64}\right)\]
\[f(x)~=~-\left( (x+{8})^2+{6} \right)\]
\[f(x)~=~-(x+{8})^2-{6}\]
- The second parabola has the same vertex \((-8,-6)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x+{8})^2-{6}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}+{16}x+{64}-{6}\]
\[g(x)~=~x^{2}+{16}x+{58}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}+{4}x-{11}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}-{4}x+{11}\right)\]
\[\left(\frac{-4}{2}\right)^2~=~4\]
\[f(x)~=~-\left(x^{2}-{4}x+{4}+{11}-{4}\right)\]
\[f(x)~=~-\left( (x-{2})^2+{7} \right)\]
\[f(x)~=~-(x-{2})^2-{7}\]
- The second parabola has the same vertex \((2,-7)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x-{2})^2-{7}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}-{4}x+{4}-{7}\]
\[g(x)~=~x^{2}-{4}x-{3}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}+{16}x-{68}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}-{16}x+{68}\right)\]
\[\left(\frac{-16}{2}\right)^2~=~64\]
\[f(x)~=~-\left(x^{2}-{16}x+{64}+{68}-{64}\right)\]
\[f(x)~=~-\left( (x-{8})^2+{4} \right)\]
\[f(x)~=~-(x-{8})^2-{4}\]
- The second parabola has the same vertex \((8,-4)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x-{8})^2-{4}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}-{16}x+{64}-{4}\]
\[g(x)~=~x^{2}-{16}x+{60}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}+{14}x-{57}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}-{14}x+{57}\right)\]
\[\left(\frac{-14}{2}\right)^2~=~49\]
\[f(x)~=~-\left(x^{2}-{14}x+{49}+{57}-{49}\right)\]
\[f(x)~=~-\left( (x-{7})^2+{8} \right)\]
\[f(x)~=~-(x-{7})^2-{8}\]
- The second parabola has the same vertex \((7,-8)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x-{7})^2-{8}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}-{14}x+{49}-{8}\]
\[g(x)~=~x^{2}-{14}x+{41}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}+{4}x\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}-{4}x\right)\]
\[\left(\frac{-4}{2}\right)^2~=~4\]
\[f(x)~=~-\left(x^{2}-{4}x+{4}-{4}\right)\]
\[f(x)~=~-\left( (x-{2})^2-{4} \right)\]
\[f(x)~=~-(x-{2})^2+{4}\]
- The second parabola has the same vertex \((2,4)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x-{2})^2+{4}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}-{4}x+{4}+{4}\]
\[g(x)~=~x^{2}-{4}x+{8}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}-{6}x-{3}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}+{6}x+{3}\right)\]
\[\left(\frac{6}{2}\right)^2~=~9\]
\[f(x)~=~-\left(x^{2}+{6}x+{9}+{3}-{9}\right)\]
\[f(x)~=~-\left( (x+{3})^2-{6} \right)\]
\[f(x)~=~-(x+{3})^2+{6}\]
- The second parabola has the same vertex \((-3,6)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x+{3})^2+{6}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}+{6}x+{9}+{6}\]
\[g(x)~=~x^{2}+{6}x+{15}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}+{6}x-{5}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}-{6}x+{5}\right)\]
\[\left(\frac{-6}{2}\right)^2~=~9\]
\[f(x)~=~-\left(x^{2}-{6}x+{9}+{5}-{9}\right)\]
\[f(x)~=~-\left( (x-{3})^2-{4} \right)\]
\[f(x)~=~-(x-{3})^2+{4}\]
- The second parabola has the same vertex \((3,4)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x-{3})^2+{4}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}-{6}x+{9}+{4}\]
\[g(x)~=~x^{2}-{6}x+{13}\]
A downward-facing parabola \(f\) is defined below.
\[f(x)~=~-x^{2}-{10}x-{23}\]
An upwards-facing parabola, \(g\), is constructed such that \(g\) and \(f\) are mirror images with a shared vertex, as shown in the graph below.
The second parabola, \(g\), can be written in standard form with two parameters (\(B\) and \(C\)).
\[g(x)~=~x^2+Bx+C\]
Find \(B\) and \(C\).
- \(B~=~\)
- \(C~=~\)
- Find the vertex by completing the square. (First factor out the negative to get a quadratic polynomial with a +1 leading coefficient.)
\[f(x)~=~-\left(x^{2}+{10}x+{23}\right)\]
\[\left(\frac{10}{2}\right)^2~=~25\]
\[f(x)~=~-\left(x^{2}+{10}x+{25}+{23}-{25}\right)\]
\[f(x)~=~-\left( (x+{5})^2-{2} \right)\]
\[f(x)~=~-(x+{5})^2+{2}\]
- The second parabola has the same vertex \((-5,2)\), but it faces upwards. The general vertex form is \(A(x-h)^2+k\). If \(A\) is positive, the parabola faces upward, and if \(A\) is negative, the parabola faces downward. If \(|A|>1\), then the parabola gets stretched vertically. If \(|A|<1\), then the parabola gets squished vertically. Anyway, in this case, we just need to switch the sign of \(A\).
\[g(x)~=~(x+{5})^2+{2}\]
- Expand to produce standard form.
\[g(x)~=~x^{2}+{10}x+{25}+{2}\]
\[g(x)~=~x^{2}+{10}x+{27}\]
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) -1
- \(h=\) -3
- \(k=\) animated
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) -1
- \(h=\) animated
- \(k=\) 5
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) 1
- \(h=\) animated
- \(k=\) 4
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) 1
- \(h=\) animated
- \(k=\) -5
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) animated
- \(h=\) -5
- \(k=\) -1
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) -1
- \(h=\) 4
- \(k=\) animated
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) -1
- \(h=\) 4
- \(k=\) animated
Try it out in Desmos!
In Desmos graphing calculator, define \(f(x)\) with the following parameterized equation, and add sliders for all three parameters (\(A\), \(h\), and \(k\)). The sliders should default with boundaries of \(-10\) and \(10\).
\[f(x)~=~A(x-h)^2+k\]
To make the animation shown below, two of the parameters were set to constants, and the other parameter was animated by pressing the "play" button. That animated parameter varies between \(-10\) and \(10\) by default.
Find the value of each parameter, or if the parameter varies then choose "animated".
- \(A~=~\)
- \(h~=~\)
- \(k~=~\)
- \(A=\) animated
- \(h=\) -5
- \(k=\) -1
Try it out in Desmos!